Find the equation of the tangent plane and symmetric equation for the normal line to the given surface at P:

xy^2 + zy^2 + 4y -xz^2 = 18
P(-2,0,3)

Define the surface as:

f(x,y,z)=xy^2+zy^2+4y-xz^2-18=0

Then check that P0(-2,0,3) is on the surface by evaluating f(-2,0,3) and confirm that it is zero.

Let fx,fy,fz be the partial derivatives of f(x,y,z) with respect to x,y and z respectively, and evaluated at P0(-2,0,3).

The tangent plane at P is then given by:
fx(x-x0)+fy(y-y0)+fz(z-z0)=0
where (x0,y0,z0) is (-2,0,3).

To find the equation of the tangent plane to the given surface at point P(-2, 0, 3), we first need to find the partial derivatives of the surface equation with respect to x and y.

The surface equation is: xy^2 + zy^2 + 4y - xz^2 = 18

Partial derivative with respect to x:
∂/∂x (xy^2 + zy^2 + 4y - xz^2) = y^2 - z^2

Partial derivative with respect to y:
∂/∂y (xy^2 + zy^2 + 4y - xz^2) = 2xy + 2zy + 4

Now, we can find the normal vector to the surface at point P by evaluating the partial derivatives at P:

∂/∂x (xy^2 + zy^2 + 4y - xz^2) = y^2 - z^2 = 0
∂/∂y (xy^2 + zy^2 + 4y - xz^2) = 2xy + 2zy + 4 = 0

Substituting the coordinates of point P (-2, 0, 3) into the partial derivative equations:

(0)^2 - (3)^2 = -9
2(-2)(0) + 2(0)(3) + 4 = 4

Therefore, the normal vector to the surface at point P is N = <-9, 4>.

The equation of a plane is given by the formula: Ax + By + Cz = D, where A, B, and C are the coefficients of x, y, and z respectively, and D is a constant. We can use this formula, along with the normal vector, to find the equation of the tangent plane.

Plugging in the values from point P and the normal vector:

-9(x + 2) + 4(y - 0) = 0
-9x - 18 + 4y = 0
-9x + 4y = 18

So, the equation of the tangent plane to the surface at point P(-2, 0, 3) is -9x + 4y = 18.

To find the symmetric equation for the normal line, we can use the point-normal form of a line equation, which is given by:

(x - x0)/a = (y - y0)/b = (z - z0)/c, where (x0, y0, z0) is a point on the line and (a, b, c) is the direction vector (the coefficients of x, y, z respectively) for the line.

For the normal line to the surface, the direction vector is the same as the normal vector N = <-9, 4>. Plugging in the coordinates of point P (-2, 0, 3):

(x - (-2))/-9 = (y - 0)/4 = (z - 3)/0

Simplifying:

(x + 2)/-9 = y/4

So, the symmetric equation for the normal line to the surface at point P(-2, 0, 3) is (x + 2)/-9 = y/4.

To find the equation of the tangent plane and the symmetric equation for the normal line to the given surface at point P(-2,0,3), we will proceed through the following steps:

Step 1: Calculate the partial derivatives
Step 2: Determine the normal vector at point P
Step 3: Write the equation of the tangent plane
Step 4: Derive the symmetric equation for the normal line

Now, let's go through each step in detail:

Step 1: Calculate the partial derivatives
The equation of the surface is given by xy^2 + zy^2 + 4y - xz^2 = 18. To find the partial derivatives, we differentiate the equation with respect to each variable:

∂f/∂x: y^2 - z^2
∂f/∂y: 2xy + 4z + 4
∂f/∂z: 2zy - 2xz

Step 2: Determine the normal vector at point P
Substitute the values of x, y, and z into the partial derivatives to get the directional vector of the tangent plane at point P(-2, 0, 3):

∂f/∂x at P(-2,0,3) = (0)^2 - (3)^2 = -9
∂f/∂y at P(-2,0,3) = 2(-2)(0) + 4(3) + 4 = 16
∂f/∂z at P(-2,0,3) = 2(0)(0) - 2(-2)(3) = 12

Therefore, the normal vector at point P is N = (-9, 16, 12).

Step 3: Write the equation of the tangent plane
The equation of a plane is given by Ax + By + Cz = D, where A, B, C are the components of the normal vector (N = (A, B, C)) and D is a constant. By substituting the values of the normal vector and the point P(-2, 0, 3) into the equation, we get:

-9(x-(-2)) + 16(y-0) + 12(z-3) = 0

Simplifying the equation, we have:

-9x + 18 + 16y + 36 + 12z - 36 = 0
-9x + 16y + 12z + 18 = 0

Therefore, the equation of the tangent plane at point P(-2, 0, 3) is -9x + 16y + 12z + 18 = 0.

Step 4: Derive the symmetric equation for the normal line
To find the symmetric equation for the normal line, we need a point on the line and the direction vector of the line. We already have the point P(-2, 0, 3), and the direction vector is the normal vector we obtained earlier, N = (-9, 16, 12).

Using the symmetric equation for a line, the equation becomes:

(x - (-2)) / (-9) = (y - 0) / 16 = (z - 3) / 12

Simplifying the equation, we have:

(x + 2) / (-9) = y / 16 = (z - 3) / 12

This is the symmetric equation for the normal line to the given surface at point P(-2, 0, 3).